Let $E$ Lebesgue measurable of finite measure. Compute $$\lim_{n\to\infty }\int_E\sin^n(x)dx.$$
I already have the solution, but I did differently, and I would like to know if it's correct or not.
We can see that $\sin^n(x)\longrightarrow 0$ pointwise.
Let $\varepsilon>0$. By Egoroff theorem there is a closed set $F\subset E$ s.t. $m(E\backslash F)<\varepsilon$ and $\sin^n(x)\to 0$ uniformly. Therefore, $$\int_E|\sin^n(x)|dx\leq \int_{F}|\sin^n(x)|dx+\int_{E\backslash F}dx=m(E\backslash F)+\int_{F}|\sin^n(x)|dx\underset{n\to\infty }{\longrightarrow} m(E\backslash F)<\varepsilon$$ Therefore $$\lim_{n\to\infty }\int_E\sin^n(x)dx=0.$$
Do you think it work ?
Yes, this is a fine proof (modulo the fact that when you write pointwise, it really should be "pointwise almost everywhere"). It can be simplified significantly, though: Use the dominated convergence theorem with dominating function $\chi_E$.
It's also worth remarking that a slight reworking of your proof gives a proof of the dominated convergence theorem (on spaces of finite measure) from Egoroff's theorem; some knowledge about continuity of measures is needed to control integral over $F$, but it's not too much extra effort.